When does trace and determinant of a 2 x 2 matrix equal each other? (Linear Algebra)
- Background Information:
I am new to linear algebra, and I recently came across this homework question that I am confused about. I appreciate any explanation that can help me improve my solution.
- Question:
What condition on the entries of a 2x2 matrix A means Tr(A) = det(A)? Provide two distinct examples of 2x2 matrices which satisfy this.
- My approach (Not Complete):
Considering the following 2 x 2 matrix, the det(A) = 4, and Tr(A) = 4
\begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}
However, considering this 2 x 2 matrix, the det(A) = 9, and Tr(A) = 6
\begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}
I think the condition would be having 2 x 2 matrix such that the matrix is (symmetric) and (n = 2).
\begin{bmatrix} n & 0\\ 0 & n \end{bmatrix}
My solution makes sense, but I feel it is incomplete. Am I missing a key point or a concept that I can add to my answer?
- Edited:
I have tried this solution with so many numbers and it seems to work. Would this be an acceptable solution?
\begin{bmatrix} a & b\\ c & d \end{bmatrix}
such that a = c = d and b = c - 2, so here is an example
\begin{bmatrix} 5 & 3\\ 5 & 5 \end{bmatrix}
det(A) = 25 - 15 = 10 , and Tr(A) = 5 + 5 = 10
$\endgroup$ 74 Answers
$\begingroup$consider below matrix $$M= \begin{bmatrix} a & b \\ c & d\\ \end{bmatrix} $$ the trace will be: $$Tr(M)=a+d$$
and the determinant : $$det(m)=(ad-bc)$$
then according to your problem:
$$a+d=ad-bc$$
so chose a and d arbitrary and then chose b and c in the way that the above equation will hold. for example:
$$a=10,d=20 $$
bc=170 and you can choose: $$ b=17 ,c=10$$
$\endgroup$ 2 $\begingroup$The condition for the matrix $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ is $a+d=ad-bc$, which becomes, for $a\ne1$, $$ d=\frac{a+bc}{a-1} $$ If $a=1$, the condition is $bc=-1$.
Thus you get a family: $$ \begin{bmatrix} 1 & t \\ -1/t & d \end{bmatrix} $$ with arbitrary $t\ne0$ and $d$.
Also \begin{bmatrix} a & b \\ c & (a+bc)/(a-1) \end{bmatrix} for arbitrary $a\ne 1$, $b$ and $c$.
Note that $d=1$, in the second case, implies $bc=-1$, so it is symmetric with the case $a=1$.
$\endgroup$ 1 $\begingroup$If the 2 dimensional matrix $A$ has 2 eigenvalues $x$ and $y$, then $Tr(A)=x+y$ and $det(A)=xy$. So then we have the equation $x+y=xy$ which transforms to:
$$ y = \frac{x}{x-1} $$
Set some value of $x$, and you'll get a value of $y$. You should then be able to choose any two linearly independent eigenvectors you want. For example, we can choose $x=y=2$, and eigenvectors $\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}$ and then we get your solution of $\begin{pmatrix}2 & 0\\0 & 2\end{pmatrix}$. Another interesting solution is $x=y=0$, which gives the zero matrix. In most cases, however $x$ and $y$ will be unequal, so we might have $x=3, y=3/2$ or $x=-1, y=\frac{1}{2}$.
And the eigenvectors can be as interesting as you like. (As long as they are linearly independent.)
EDIT: As was kindly pointed out to me, to be able to cover non-diagonalizable matrices we need to also allow generalized eigenvectors.
$\endgroup$ 4 $\begingroup$Consider an arbitrary $2\times2$ matrix, $$ A= \begin{bmatrix} a & b\\ c & d\\ \end{bmatrix} $$ Its determinant is given by $$ \det(A)=ad-bc $$ and its trace is given by $$ \text{Tr}(A)=a+d $$ So, we want to know when $\det(A)=\text{Tr}(A)$. That is, when is $$ ad-bc=a+d $$ See if you can figure out the conditions on the entries that make the left hand side equal the right hand side.
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